Answer
See the detailed answer below.
Work Step by Step
Just after closing the switch, the current is $I_{max}$ and decreases until it reaches zero when the capacitor is fully charged and $Q=Q_{\rm max}$.
To analyze the circuit, we can use Kirchhoff's loop law. Starting from the left lower corner and moving clockwise,
$$\varepsilon-IR-\dfrac{Q}{C}=0$$
Recalling that $I=dQ/dt$,
$$\varepsilon-\dfrac{dQ}{dt}R-\dfrac{Q}{C}=0$$
$$\varepsilon-\dfrac{Q}{C}= \dfrac{dQ}{dt}R$$
$$dQ=\dfrac{\varepsilon-\dfrac{Q}{C}}{R}dt$$
$$dQ=\dfrac{ \varepsilon C-Q }{CR}dt$$
$$CRdQ=( \varepsilon C-Q )dt$$
Hence,
$$\dfrac{dQ}{ \varepsilon C-Q}=\dfrac{dt}{CR}$$
Integrating both sides:
$$\int_0^{Q }\dfrac{dQ}{ \varepsilon C-Q}=\int_0^t\dfrac{dt}{CR}$$
$$-\ln(\varepsilon C-Q)\bigg|_0^Q= \dfrac{t}{CR}\bigg|_0^t$$
$$-\ln(\varepsilon C-Q) +\ln(\varepsilon C)= \dfrac{t}{CR} $$
$$ \ln(\varepsilon C-Q) -\ln(\varepsilon C)= -\dfrac{t}{CR} $$
Hence,
$$\ln\left[ \dfrac{\varepsilon C-Q}{\varepsilon C}\right]= -\dfrac{t}{CR} $$
Thus,
$$ \dfrac{\varepsilon C-Q}{\varepsilon C} = e^{-t/CR }$$
$$ \varepsilon C-Q =\varepsilon C e^{-t/CR }$$
Therefore,
$$Q= \varepsilon C-\varepsilon C e^{-t/CR }$$
$$Q= \varepsilon C(1-e^{-t/CR })$$
where $RC=\tau$, and $Q_{\rm max}= \varepsilon C$,
$$\boxed{Q= Q_{\rm max}(1-e^{-t/\tau})}$$
